How do you simplify the expression #(y/x-x/y)/((x+y)/(xy))#?

Question

14750 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-05T05:09:19

#(y/x-x/y)/((x+y)/(xy))=y-x#

#(y/x-x/y)/((x+y)/(xy))#

= #((y^2-x^2)/(xy))/((x+y)/(xy))#

=#(y^2-x^2)/(xy)div(x+y)/(xy)#

= #(y^2-x^2)/(xy)xx(xy)/(x+y)#

= #((y+x)(y-x))/(cancel(xy))xx(cancel(xy))/(x+y)#

= #((x+y)(y-x))/((x+y))#

= #(cancel((x+y))(y-x))/(cancel((x+y)))#

= #y-x#